MENSURATIONMENSURATION, or the art of measuring, involves the construction of measures, the methods of using them, and the investigation of rules by which magnitudes which it may be difficult or impossible to measure directly are calculated from the ascertained value of some associated magnitude. It is usual, however, to employ the term mensuration in the last of these senses; and we may therefore define it to be that department of mathematical science by which the various dimensions of bodies are calculated from the simplest possible measurements.
The determination of the lengths and directions of straight lines, including what are familiarly known as problems in heights and distances, generally depends on the solution of triangles, and will be discussed in the articles Trigonometry and Surveying. The remaining portions of the subject are the determinations of the lengths of curves, the areas of plane or other figures, and the volumes and surfaces of solids; and it is of mensuration as thus restricted that the present article will discuss some of the more important problems.
§ 1. Units of Length, Area, and Volume.— In measuring any magnitude we select some standard or “unit” to measure by. Thus in measuring length we take for unit an inch, a foot, or a yard. From the unit of length we derive the units of area and volume. Thus we define the unit of area to be the area of the square described upon the unit of length, and the unit of volume to be the volume of the cube whose edge is the unit of length or whose side is the unit of area. For example, if an inch be taken as the unit of length, the square whose side is 1 inch is the unit of area, and the cube whose edge is 1 inch is the unit of volume. The length of a line, the area of a surface, and the volume of a solid are then expressed by the numbers, whole or fractional, of units of length, area, and volume which they respectively contain. Hence, if I denote the linear unit, the length of a line which contains a units is al, or simply a since I is unity; similarly the area of a surface which contains b units of area is bm, or simply b, where m is the unit of area.
§ 2. Commensurable and Incommensurable Magnitudes.— When two magnitudes have a common measure, that is, when another magnitude can be found which is contained in each an exact number of times, they are said to be “commensurable.” Thus a line 4⅜ and another 3⅛ inches long are commensurable; for, if ⅜ inch be taken as unit of length, the former contains the unit nine times and the latter seven times. If no common measure can be found, the two magnitudes are said to be “incommensurable.” For instance, 1 and √2 have no common measure; for V2 = 1'4142 . . . an interminable decimal, and hence no unit, however small, can be found which will be contained in each an exact number of times. If, however, we take χ∕2 = Γ4, the error will be less than τ ⅛j if V2 = 1'414, the error will be less than τj ⅛ π , &c. Hence, by taking a sufficient number of figures, we can find a fraction which will differ from χ∕2 by less than any assignable quantity, and therefore we can always find two commensurable magnitudes that will represent two incommensurable ones to any degree of accuracy we please. In what follows we need therefore only consider commensurable lines.
§ 3. Area of a Rectangle.— Let the side AB (fig. 1) contain a units and the side BC b units of length. If we divide AB into a equal parts, each equal to the unit of length, and similarly BC into b equal parts, and if through the points of division we draw lines parallel to the sides of the rectangle, these lines will divide the rectangle into a series of rectangles, each of which is the unit of area, since each is a square whose sides are of unit length. As we have a rows of these rectangles, and b in each row, the whole number of rectangles will be ab. Therefore area of ABCD= ab units of area *=ab.
PART I.—PLANE FIGURES.
Section I.— Plane Figures contained by Straight Lines.
A. The Rectangle.
§ 4. Let ABCD (fig. 2) be a rectangle, and let AB = CD=α, BC = DA = 5, AC = c, and the angle BAC = a ; it is required to find its area. Since a rectangle is completely determined when two independent data, one of which at least is a length, are given connecting its parts, we can determine its area in the following cases.
(a) When its length a and its breadth b are given. — It has already been proved (§ 3) that area of ABCD = ab ; or the area of a rectangle is equal to its length multiplied by its breadth.
Example.— Let α = 12 feet 6 inches and δ=9 inches, then area of ABCD = 12'5 × '75 = 9'375 square feet.
If we make use of logarithms in the above calculation we have log area=logα + logδ. logα = logl2'5 =1'0969100 log b = log '75 = 1'8750613
therefore log area = '9719713; hence area-= 9 '375.
(3) When a side a and the diagonal c are given.— By Euclid i. 47 we have
5 2 = c 2 -α 2 , or δ = Vc 2 -a 2 , therefore area of ABCD=αδ=αVc 2 -α 2 5 or logarea=loga + ⅜log(c + a) + ∣log(c-a).
Example.— Let a=238 and c=456, then loga-log238 = 2'3765770 ⅛log(c + a) = ⅜log694 = 1 '4206797 ⅛ log(c - a) - ⅛log218 = 1'1692282
therefore log area = 4 '9664849; hence area = 925/3'1.
(γ) When a side a and a its inclination to the diagonal are given. —Since = tan<x, 5 = αtanα, a and therefore area of ABCD = ab =α 2 tan a; or logarea=21ogα + Ltanα-10.
Example. — Let a = 36 and a = 32° 25' 15", then
21oga = 21og36= 3'1126050
Ltana=Ltan32° 25' 15" = 9'8028622
therefore logarea=12'9154672-10 = 2'9154672; hence area = 823'127.
(5) When the diagonal c and a its inclination to either of the sides are given. — W e have
α = ccosα, andδ=csina, therefore area of ABCD = ab =c 2 sin αoosa=⅛c 2 sin 2a; or 2area=c 2 sin2a, and hence log2 area=2loge+ Lsin2α -10.
§ 5. A square being a rectangle whose sides are equal, we can at once determine its area. When one datum, which must be a length, is given the square is completely determined, and hence we have only two cases to consider.
(a) When the side is given.— From § 4, a, we have at once area of square = ab =a × a=a?.
(3) When the diagonal c is given.— From § 4, 3, we have
α 2 + α 2 = c 2 , or a 2 = ⅜c 2 ; hence area of square=α 2 = ⅛c 2 , or 2 area = c 2 ; and therefore log 2 area = 2 log c.
B. Right-angled Triangles.
§ 6. The diagonal of every rectangle divides it into two equivalent right-angled triangles (Eucl. i. 34), and hence the area of the right-angled triangle ABC (fig. 2) is equal to half the area of the corresponding rectangle ABCD.
C. Triangles Generally.
§ 7. In every triangle there are six elements to be considered, namely, the three sides and the three angles. If any three of these six be given, provided one is a length, the triangle is completely determined, and hence its area can be found.
§ 8. Length of Perpendiculars of a Triangle.— In the triangle ABC (fig. 3) let BC = α, CA = 5, AB=c, AD the perpendicular from A on BC=⅛, BD=ae, and CD = y.
Since BDA and CDA are right angles, we have c 2 =aj 2 + 7ι 2 , and ⅛ 2 = y 2 + Λ 2 , and therefore
5 2 - c 2 =y 2 - aj 2 =(y+»)(y - x) ≈a(y - x);
δ 2 -c 2 whence y-x≈ . Cl
But y ÷α!=α, and, by solving these equations, we obtain
δ 2 + α 2 -c 2 y~ 2α -·
Again h*-V-y*≈V - ^ a +^ 2 -<∙ s y W-(⅛ 2 + α*-c 2 ) 2
(α + ⅜ + c)(⅛ + c - α)(c + « - b)(a + b-c )
4α 2 hence Λ=-A-√(α+5 + c)(5 + c-α)(c + α-δ)(α + b -c).
2α
Now let a + b + c≈2s, then 5 + c-α = 2(s-α), c + α-δ = 2(s-δ), and α + 5-c = 2(s-c).
Therefore, on substituting and reducing, we obtain
2
h = — √s(s - a)(s - b)(s - c).
Similarly the perpendiculars from B and C on the opposite sides are respectively
ο 2
y√s(s-α)(s-J)(s-c) , and —√i(s-α)(s-φ-c).
§ 9. We now proceed to investigate formulae for the area of a triangle in the following important cases.
(a) When the base a and the altitude h are given.— Since a triangle is equel to half a rectangle of the same base and altitude, we have at once area ABC —⅛α⅛.
Example.— Let α = 40 chains and A-14'52 chains, then area - ⅜ × 40 × 14 '52 = 290'4 square chains.
(3) When two sides a and c and the included angle B are given.— From fig. 3 — — sinB , and therefore Λ=csinB; hence area— ⅛ah- ⅛αcsinB; or log2area=logα + logc + LsinB-10.
Example.— Let a —40, c—30, and B —30°, then area = ⅛ac sin B — ⅜×40×30×⅛- 300.
(γ) When the three sides a, b, c are given.— From § 8
⅛ = - Vs(s-α)(s-δ)(s-c) , and therefore area = ⅛ah - ⅜α × -∣- √s(s - a)(s - b)(s - c) - Vs(s - α)(s - b){s - c); or log area — ⅛ {logs + log (s - α) + log (s - b) + log (s - c)}.
Since 2s=a + b+c, we have area of triangle — ∣ V2(α 2 δ 2 + b i c i + c 2 a 2 ) - (a i + b i +c 4 ).
Example 1.—Leta = 13, 5 = 14, and c—15, then s-⅜(13 + 14 + 15)-21, s-α-21-13 = 8, s-b≈ 21-14 — 7, and s-c—21-15=6; therefore area = √21 × 8 × 7 × 6 = 84.
Example 2.—Let a —255, δ = 238, and c = 221, then logs-log 357= 2'5526682
log (s-α) = log 102— 2'0086002 log(s-δ) = logll9= 2Ό755470 log (s - c) = log 136 = 2Ί335389
therefore logarea —J(8'7703543) = ^'3851771; hence area = 24276.
(δ) When any two angles B and C and the adjacent side a are given. —Since c sinC αsinC
a sinA ’ sinA ’ and therefore (by 3)
area = ⅛αcsinB - ? 8smBsln Ç where A = 180°- (B + C),
2 sin A
or log 2 area=2 log a + Lsin B + I. sin C + I. cosec A - 30.
Since all the angles of a triangle are given when any two are given, we can find the area of a triangle when any two angles and any one side are given. Thus, when A, B, and c are given, we know C also, and the problem reduces to a case of the preceding.
(«) When the three medians a, β, y are given.— If a, b, c be the three sides of a triangle, and α, 3, y the three medians, i.e., the lines drawn from the angles to the middle points of the opposite sides, then by well-known geometrical propositions we have
4(α 2 + 3 2 + τ 2 ) = 3(α 2 + ⅛ 2 + c 2 ),
16(a 2 3 2 + 3 2 τ 2 + τ 2 a 2 ) = 9(a⅞ 2 + δ 2 c 2 + c 2 a 2 ), and 16(a 4 + 3 4 + γ 4 ) = 9(a 4 + 5 4 + c 4 ).
Now (§ 9, γ)
area of triangle — ∣ V2(α 2 6 2 + 5 2 c 2 + c 2 α 2 ) - (a* + b* + c 4 ), therefore = ⅛√2(α 2 j8 2 + 3 2 γ 2 + 7 2 a 2 ) - (a 4 + 3 4 + γ 4 ).
D. Parallelograms.
§ 10. The opposite sides and angles of a parallelogram being equal, three independent data, one of which at least is a length, are necessary and sufficient to determine it completely.
In theparallelogramABCD (fig. 4) let BC = DA = α, AB = CD = δ, AC=c, AE = ⅛, the angle ABC = α and AOD = 3∙
Since the diagonal AC divides the parallelogram into two equivalent triangles, we obtain
(a) area of ABCD —2 area of triangle ABC
= 2X⅛ a×h (§9, a)=ah;
(3) area of ABCD — 2area ABC-2 × ⅛α⅛sinα (§ 9, 3) = a5sina; or logarea = loga + logδ + Lsina - 10;
(7) area of ABCD = 2 area ABC = 2(ABO+ CBO)
= 2{⅛BO. A0sinA0B + ⅜BO.COsinCOB}-2{⅜BO.ACsin3}
= ⅜BD. AC sin 3 = ⅜c<7sin 3 , or Iog2area=logc + Iog<i + Lsin3-10.
§ 11. If the parallelogram be equiangular (a rectangle), c— d, and area — ⅜c 2 sin3∙ If it be equilateral (a rhombus), 3 = 90°, and area =⅜crf. If it be both equiangular and equilateral (a square), c— d and 3 = 90°, and area — ⅛c 2 as before (§ 5, 3).
E. Trapeziums.
§ 12. To determine a trapezium completely four data are necessary and sufficient.
In the trapezium ABCD (fig. 5)letBC = a, CD — b, DA=c, AB-ci, and AE perpendicular to BC — h, and draw AF parallel to CD, then
(a) area ABCD —area ABC + area ADC
= ⅛α⅛ + ⅛ch ≈⅛(a + c)h; or the area is equal to half the sum of the parallel sides multiplied by the perpendicular between them.
Again, area of ABF -⅜BF × AE (§ 9, α) = ⅜(α- c)h, also area of ABF = Vs(s-AB)(s-BF)(a-FA), where 2s—AB + BF + FA; hence Λ=^-^√s(s-AB)(s-BF)(s-FA), therefore
(3) area of ABCD = ⅛(α + c)A=^√ s (s - AB)(s - BF)(s - FA)
(- α +δ+ c + i ≈0(α ÷δ-c-ti)(α + δ-c + ci)(α-5-c + ci) l since AB=ci, BF— a-c, and FA = CD — b.
Thus we can find the area of a trapezium in terms of its sides.
§ 13. If c—0, ABCD becomes a triangle, and its area
= ∣∖∕( -α+5+ci)(α+δ-ci)(α + ⅛ + d)(α-5 + ci).
Again, if c=α, then also b=d, and ABCD becomes a parallelogram, and its area takes the indeterminate form-^-, as it should do, since four sides do not completely determine a parallelogram.
F. Quadrilaterals Generally.
§ 14. A quadrilateral is completely determined when five independent data are given. We consider the following cases.
(a) When any diagonal and the perpendiculars on it from the opposite vertices are given.
The quadrilateral ABCD (fig. 6) = ABD + BCD
= ⅛BD.AE + ⅜BD.CF
= ∣BD(AE + CF)j or the area is equal to half the product of the diagonal and the sum of the perpendiculars.
If the diagonal BD fall without the figure, as in the concave quadrilateral ABCD (fig. 7), then it is clear that area ABCD = ⅛BD( AE - CF).
(3) When the diagonals and their included angle are given.— In the quadrilateral ABCD (fig. 8, p. 16) let BD — h, AC = h, and angle DEA — a, then
ABCD-ABD+BCD
= ⅛BD.AEsinα + iBD.CEsinα (§ 10, 7)
= ⅛Λ(AE + CE)sina
— ⅜λλ'sina; or the area is equal to the product of the diagonals and the sine of their contained angle.
The same result holds when one of the diagonals falls without the quadrilateral, as in fig. 7, as the reader can easily verify.
(7) When the four sides and the angle between the diagonals arc given.— If a, b, c, d be the sides and a the angle between the diagonals it can easily be shown that area of quadrilateral = ∣(α 2 - b 2 + c i - cP) tan a.
(δ) When the four sides are given and the opposite angles are supplementary.— In fig. 8 let AB = α BC = δ, CD—c, DA—d, AC = ½, angle ABC = α, angle CDA = 0, and let α + 3 = 180°, then area of ABCD = ABC + ADC
= ⅛ab sine + ⅜c<i sin β.
But sin j8 — sin (180°- α)—sin α, therefore area of ABCD = ⅛(ab + cd) sin α.
This gives us the area of the quadrilateral in terms of the four sides and one angle.
Again we have
α 2 + ⅛ 2 - 2ab cos a=h? ≈ c 2 + cP - 2cd cos β = c 2 + cf 2 + 2cd cos a,
., , α 2 + ⅛ 2 -c 2 -d 2 , , therefore cosα= —· .-. - κ — , and hence
2(αo + cd)
(α + δ + c-cf)(α + 5-c + d) , l + cosα∙=- ,'∖ ' and
2(α⅛ + cd) ’ 1 (c + d+α-δ)(c+<Z-α + δ)
X I.·UOU — — ·
2(ab + cd)
From this we obtain sin 2 α= (1 +cosa)(l - cos a)
(δ + c + d-a)(c + d + a-δ)(rf + a + ⅛-c)(a + δ + c-<Z) 4(ab+cd)*
Now let a + b+c+d=2s, then ⅜ (ab + cd) sin a = √(s - a)(s -b)(sc)(s - d); therefore area of ABCD = V(s - α)(s - b)(s - c)(s - d), or log area = ∣{log(s-α) + log(s-δ) + log(s-c) + log(s-iZ)} .
If d—0, the qua drilateral becomes a triangle, and its area is V (s - a)(s - b)(s - c)s as before. In extracting the square root of sin a we take the positive sign, since the angle a is less than two right angles.
G. Regular Polygons.
§ 15. Since a regular polygon is both equilateral and equiangular, a circle can be inscribed within it and also described about it, and thus the n straight lines drawn from the common centre of the two circles to the n vertices of the polygon divide it into n triangles equal in every respect. Therefore the area of the polygon is equal to n times the area of any one of these triangles.
§ 16. Radius of Inscribed and Circumscribed Circles.— Let AB (fig. 9)=αbea side of a regular polygon of n sides; let C be the centre of the inscribed and circumscribed circles, CD=r the radius of the former, and CE=R the radius of the latter. The angle ACB is evidently equal to the nth part of four right angles, that is
ACB = - , and ACD = ⅛ACB = 18 -. η n
Now AD=-^--CDtanACD = rtanl^-,
2 n and AD = -^- = ACsin ACD = Rsinl^-;
2 n therefore r=α×icot- , n and R=αx ⅜ cosec— θ.
n
§ 17. Perimeter of Polygon.— The perimeter of the polygon of 180° Ί R∏ o n sides is na, i,e., 2 nr tan f or 2πRsin-—
η n
From this it follows that the perimeters of the inscribed and circumscribed regular polygons of n sides of a circle of radius r are o · 180 ° 4 « t 180° i . 1
2nrsιn-^- and 2nrtan - — respectively.
§ 18. Area of Polygon.
(a) In terms of r.—The area of polygon = nACB≈nAD.CD = n× r 2 tanl 8 θl .
n (β) In terms ofR.— The triangle ACB
= ⅜AC. CBsinACB = ⅛R 2 sinθθθl, n and therefore area of polygon ≈∣nR 2 sin -θθ-.
(γ) In terms of a.— The triangle ACB
up nn a a i , 18° 0 α 2 180° = ⅜AB. CD = -g-×r=-× ⅛αcot-—ψcot~^- , and therefore area of polygon =α 2 × — cot^^
4 n or log4area=logn + Lcotl^- + 21ogα-10.
n
From α and β it follows that the areas of the inscribed and circumscribed regular polygons of n sides of a circle of radius r are 360° 180°
⅜nr 2 sin —— and nr 2 tan-—respectively.
§ 19. In the formula (§ 18, γ) for the area of a polygon, the factor ∣cot ι 8 θ - has a definite value for every value of n, and hence, if we find its value once for all for a large number of values of n, and tabulate the results, we can find the area of a regular polygon of n sides by multiplying the square of its side by the appropriate tabular value.
Again, if α = l we have r=¾coti^- and R =⅜cosec^-! η n ’ and thus we obtain in a similar manner the radius of the inscribed and circumscribed circles by multiplying the side by the appro- 180° 180°
priate tabular value of ⅛cot-— and ⅛cosec-respectively.
§ 20. The following table contains the values of — cot^^- and 4 n
180° 180°
their logarithms, and the values of ⅛cot — and ⅛ cosec—— for all η n values of n from 3 to 12. [table]
Let A denote the area of a polygon of n sides and A , the corresponding tabular value of—cot ί— , then
4 n A = α 2 A', or log A = 2 log a + log A'.
H.— Length of the Radius of the In&cribcd, Escribed, and Circumscribed Circles of a Triangle.
§ 21. (a) Radius of Inscribed Circle.— LetO (fig. 10) be the centre of the circle inscribed in the triangle ABC and touching the sides in D, E, and F. Join OA, OB, and OC. The angles at D, E, F are right angles (Euch iii. 18). Let BC = α, CA = b, AB—c, and OD = OE = OF-r.
Now A B C - B O C +COA+AOB
= ⅜αr + ⅛δr + ⅛cr=⅜(α + ⅛ + c)r=rs; whence — area ^BC ∖∕s(s -a)(sb)(s -^c) ’ = s” s (β) Radius of Escribed Circles.— Let CD = OE = OF = r α , then ABC=ACO+ABO-BOC
= ⅛br a + ⅛cr a - ⅜ara = ⅜(⅛ + c-a)ra = r a (s -a),
, area of ABC ∖∕s(s - a)(s - b)(s - c) ∕s(s-b)(s-c) and rα = = __ = λ ∕ .
Similarly ∕⅜ t -< t )(,→) and rι _ ∕≠-α)(>-⅜). ∖∕ s — b ∖∕ s — c (γ) Radius of Circumscribed Circle.— Let AD (fig. 12) =7? the perpendicular from A on the side BC, and AE = 2R the diameter of the circle, then (Encl. vi. C) we have
2R ×p=b×c, therefore 2R × ap=dbc. Now ap≈2∆, where Δ denotes the area of ABC; hence R = —=—. a bc .
4∆ 4 √ s (s _ β ξ s _ jξ β _ c )
Example.— Let a = 13, δ = 14, and c=15; then r will be found to be 4, r a 10⅜, n 12, r c 14, and R 8j.
Section II.— Plane Figures contained by Curved Lines.
A. The Circle.
§22. Circumference of a Circle.— If we inscribe in any circle a regular polygon of n sides, and also circumscribe a regular polygon of the same number of sides, it is clear that the perimeter of the circle is intermediate between the perimeters of the inscribed and circumscribed polygons, and that the difference between the perimeters of the inscribed and circumscribed polygons can be made as small as we please by sufficiently increasing n. A similar statement holds with reference to the areas of the circle and the inscribed and circumscribed polygons. With the above assumptions it is easily proved that the circumference of a circle bears a constant ratio to its diameter. Hence we have
Circumference=C = constant × radius = constant × r.
It is usual to denote this constant by 2τr, and therefore
C = 2πr= πd, where d is the diameter of the circle.
§ 23. Numerical Value of τr.—The constant π being, as can be easily proved, an interminable decimal, we can only approximate to its value, but this we can do to any degree of accuracy we please.
If s and σ denote respectively a side of the inscribed and circumscribed polygons of n sides, and s' and σ' a side of the inscribed and circumscribed polygons of 2n sides, it can easily be shown that ( α ) σ= √T 2 '^- ⅛s)a > (3) s' 2 = 2r j r - √r 2 - (Js)≡ j ,
(γ)∙σ' —,
√r 2 -(⅛s') 2 where r is the radius of the circle.
If we take r=⅜ we find, by means of these formulae, and by assuming the value of s when n=6, that the perimeter of inscribed polygon of 96 sides = 3 Ί40 .... , and the perimeter of circumscribed polygon of 96 sides =3 ’ 142. . . .
From this we learn that the circumference of the circle, in this oase π, is greater than 3140 .... and less than 3142 . . . . , and therefore as far as the second place of decimals
τr=314.
By taking greater and greater values of n we obtain closer and closer approximations to τr.
The preceding method for approximating to the value of π is the simplest afforded by elementary geometry, and is also the oldest; but better and more rapid methods are furnished by the higher mathematics. The calculation of π has been carried to 707 places of decimals, the following being the first 20 figures in the result:—
314159265358979323846.
For all practical purposes it is sufficient to take
π = 314159or=⅞∣.
§ 24. The following table contains the functions of π that are of most frequent occurrence in mensuration :— [table]
§25. Units of Angular Measurement.— In measuring lines we select some line of constant length as the standard or unit; similarly in measuring angles we require to take some angle of constant magnitude as unit angle. The right angle is by its nature the simplest unit angle, but, for convenience, we take the ⅛th part of a right angle for unit, and call it a degree, which is subdivided into sixty equal parts called minutes, and these again into sixty equal parts called seconds. For theoretical purposes we define the unit angle to be the angle subtended at the centre of a circle by an arc equal to the radius. This angle we call a “radian.” In many treatises the radian measure of an angle is called the circular measure.
§ 26. The radian is a constant angle.— Let OA (fig. 13) = arc AB = r, then A0B = radian, and let AOD = 90°j then arc AD = ⅛ × 2τr= ⅞τrr; and, since angles at the centre of a circle are proportional to the arcs on which they stand (Euch vi. 33), number of degrees in radian AOB AB_ r _ 2 number of degrees in AOD AD ⅛πr π ’ therefore number of degrees in radian
2
= 90° × — = 57° ∙ 2957795 = constant.
7Γ
§ 27. Number of Radians in any Angle.— Let AOC (fig. 13) be any angle, AOB the radian, and AC=s; then number of radians in AOC AC s .
one radian AB r ’ therefore number of radians in AOC=- .
r
If AOC=90°, then s=⅛πr, and number of radians = ⅜ιr; there are thus 7r radians in two and 2π in four right angles.
When r≈l we have number of radians=s, and hence in some treatises for the number of radians in an angle we find the length of the arc given.
§ 28. To transfer from degrees to radians and conversely.— Let x denote the number of degrees in an angle, and θ the number of radians in the same, then, since »
<∙>'-⅛>
§ 29. The following table contains the values of θ for values of x up to 180°, and also for minutes and seconds. [table]
As an example of the use of this table we proceed to find the value of θ when ic=68° 45' 17"’8.
Whenx=68° 0 = 1’1868239, X = 40' θ= Ό116355, x- 5' θ= Ό014544, x =10" θ= Ό000485, x= 7" ∙0000339, and when x=O'' ’ 8 0 = ’0000039, therefore when x=68° 45' 17"’8 0 = 1’2000001.
§ 30. Combining the results of §§ 27 and 28 we obtain r ∖ n s j 180 s . v r π r W r ’.I»» θ 7Γ X
ω s . rβ .i. ra .
§ 31. Length of Arcs of Circles.— The following are the more important cases:—
(a) In terms of the chord of the arc and the radius of the circle.— Let AB (fig. 14) = 2c, AC = r, and AEB=s, then
AD = ⅛AB=c=rsin⅜C, whence C is known, and therefore the arc s is found (§ 30, γ).
(3) In terms of the height of the arc and the radius of the circle.— Let DE =h = height of arc, then
CD = CE-DE = r-Λ,
, ιr , CD r-h and cos⅛C=-—= ,
AC r whence C is found, and therefore s.
§ 32. Huygens's Approximation to the Length of a Circular Arc.— Let AB (fig. 15) ≈p be the chord of the arc AEB, and AE = EB = g ’ that of half the arc, then the arc AEB = ⅛ (85 - p) approximately.
For, let r denote the radius, s the arc AEB, and 20 the angle ACB, then 0 = ^ 7 . Again, AB=p = 2AD = 2rsin0=2rsin^-; and similarly q =2r sin — .
4r
0 3 0 5
Now sin0 = 0-∣-g-+ ∣-g--&c.; ., f n ∖ s 1 I s ∖ 8 1 I s ∖ s „ ) therefore p = 2r < r -( ττ-I + rH 1 - &c. > p I 2r ∣3^∖2r∕ ^∣5∖2r∕ )
s 8 , s 5 =S ~2 .3 . 4 . r 2 + 2 . 3 . 4 . 5 . 16 . r 4
Stately 8 f -lβr j ⅛- j ∣(⅛)∙ +l ⅜(Λ)∙ - llc . j s 8 s 5 = 4s - 2 . 3 . 4 . r 2 + 2 . 3 . 4 . 5 . 64 . r 4 ^ &C ‘
Hence, neglecting powers of — be- r yond the fourth, we obtain fcp=√ 1 _ s4 Λ= s
3 Λ 1 7680r 4 ∕ approximately.
In practice it is sometimes more convenient to use the equivalent form s = 2q + ⅛(2q-p').
§33. Area of Sector of a Circle.— Let the sector be OAB (fig. 16). Divide the arc AB into n equal parts, and draw the chords of these. Let P denote the perimeter of the broken line AB, A the area of the polygon AOB, and p the altitude of any one of the n equal triangles of which this polygon is made up.
Now in the limit, when n is indefinitely increased, P becomes the arc AB=s, a result which we symbolize thus—
T j P = arcAB=s.
j -'n=oo
Similarly I j a - a0 P “radius= r.
Again, the area of the sector is equal to the area of the polygon when the broken line AB becomes the arc AB, that is, sector=I. w=x A = I. n = αo ∖P × P = ⅛I. n ≈ 00 P × I. a=00 P · = ⅛rs.
§ 34. Let 8 denote the area of a sector of a circle, then, by means of the above result and § 27, we have
(a) S = ⅜sr, (β) S = ⅛r0 . r = ⅛r 2 0.
§ 35. We proceed to find the area of a sector of a circle in the following additional cases :—
(a) When the chord of the sector and the radius of the circle are given.— In fig. 14 let AB = 2c, and let AC = r, then we have
. ACB AD c sm ^-=-rττ=—;
2 AC r whence ACB and therefore 0 is known, and S can be found by § 34. ACB has two values, the smaller one giving the area of the minor, and the larger that of the major conjugate sector.
(3) When the chord and height of the chord are given.— Let DE (fig. 14) =Λ and AB = 2c, then
AC 2 = r 2 = AD 2 + DC 2 =c 2 + (r-Λ) 2 , whence r= c -^-, and therefore by previous case the area can be found, (γ) When the chord and angle subtended at the centre are given.
—Let AB (fig. 14) =2c and ACB = 0, then c . ACB c
— = sm — s - , or r = - , ,, r 2 sm⅛0 therefore area of sector = ⅜r 2 0=⅜ ( . c , I ×0.
∖sm⅛0∕
§ 36. Area of a Circle.— The circle being a sector whose arc is tho whole circumference we obtain at once area of circle = ⅛r × s =⅜r × 2πr = τrr 2 .
An independent proof of this proposition might be given by means of the inscribed and circumscribed polygons, and from the area of a circle the area of a sector can be deduced. The infinitesimal calculus affords a simple and elegant proof (see § 44).
§ 37. If A denote the area, r the radius, d=2r the diameter, and C the circumference of a circle, we have
(a) A = πr 3 ,
(0) A≈⅜×2πrxr=⅜Cr,
. . . 4ιr 2 r 2 C 2 ω a -ι√--s∙
(∙)
Whence we see that the area of a circle is obtained by multiplying (a) the square of its radius by π = 3Ι4159, (3) the radius by half the circumference, (γ) the square of the circumference by -I. = Ό7957, (δ) the square of the diameter by ∣π = - 78539.
§ 38. Again, from the above formulae we deduce
(a) r— A =’5641896 × A,
Vιr
(3) d —La=Γ1283792×A,
^ s ∕π
(γ) c=2√πA = 3’5449077 ×A, thus obtaining radius, diameter, and circumference from area.
§ 39. Area of a Circular Ring. — Let r and r 1 denote the radii of the outer and inner circles respect-
ively (fig. 17), then the area of the space between them
= πr 2 - πr J = π(r+r 1 )(r - r 1 ).
The circles need not be concentric, and the reader should note that the area of the ring is equal to the area of an ellipse whose major and minor axes are r + r 1 and r-r l (see § 51).
§ 40. Area of the Sector of an Annulus.— Let angle ACB = 0 in fig. 17, then the area of ABED
= sector ACB - sector DCE
= ir 2 0-⅛r 2 0, = i0(r + r 1 )(r-r 1 ).
Again, let AB — I, DE =» l 1 , and CA - CD — r - r 1 — ħ, then I , Z 1 .. .
r-- 3 - and r 1 = -±- ; therefore σ σ
r+*ι~∣∙(*+⅞λ and area of sector =≈⅛0(r + r 1 )(r-rι)-⅛0 = JA(Z + Z 1 ).
§41. Area of a Segment of aCircle.— (a) When the radius and the angle subtended at the centre are given.— In fig. 14, let AEB be a segment of a circle, then its area
— sector ACB - triangle ACB. - ⅛r 2 θ - ⅜r 2 sin θ (§§ 9, 34) — ∣r 2 (θ-sin0).
If the segment be greater than a semicircle sinβ is negative and the formula becomes
⅜r 2 (0 + sin0), as is also geometrically evident.
We might in a similar manner find the area of a segment of a circle (0) when the chord and radius are given, (γ) when the chord and its height are given, (δ) when the radius and height of the chord are given, (e) when the chord and angle subtended by the chord are given. In all these cases the method of proceeding is obvious, a segment being the difference between a sector and a triangle.
§ 42. Area of a Lune.— Let ADB and ACB (fig. 18) be two segments of circles, then the area of the lune ADBC
—segment ADB - segment ACB.
Hence if we so choose our data that we can determine the areas of the two segments we have only to take their difference to find the area of the lune.
§ 43. Area of a Circular Zone.— Let AB and CD (fig. 19) be two parallel chords, then the area of the zone ABCD
— circle - segment AHB - segment DFC; or — segment AED + trapezium ABCD + segment BGC — 2 segment AED + trapezium ABCD.
§ 44. The Infinitesimal Calculus (q. v.) furnishes a simple and elegant proof of the formulae for the areas of a circle and a sector. If y=-φ(x) be the equation to a plane curve referred to rectangular axes, then the area between the curve, the axis of x, and two ordinates corresponding to the abscissae a and b is represented by the integral
∕* i ∕ φ{x}dx. β ∕α
Let x and y be the coordinates of P (fig. 20), then if OP=r the equation to the circle is y 2 =r 2 -x 2 , and therefore area of quadrant AOB=∕ ∖∣ r t -a^dbx Jo and therefore area of whole circle — ∙nr 3 .
§ 45. If the equation to a plane curve be given in polar coordinates, the area bounded by two radii and the curve is equal to
1 f θ 2
⅜∕ , where θ 1 and 0 2 are the values of θ corresponding to the limiting radii.
For example, let AOP (fig. 20) be θ, then area of circle r 2 c 2πr λ ≈-∕ dθ≈~~πr 2 as before. ⅛∕ 0
The area of a sector can be found in a similar manner.
B. The Parabola.
§ 46. Length of an Arc of any Plane Curve.— If a plane curve be referred to rectangular axes, then the length of any arc of the curve
-√b÷(∣∏ i -∕{ ι ÷(l)!⅛
taken between proper limits, i,e., the extremities of the arc. See Infinitesimal Calculus.
§ 47. Arc of a Parabola.— Let the axes of coordinates be the axis of « and the tangent at the vertex A (fig. 21), then, the equation to the parabola being y 2 =2?rø, where m = 2α = ⅜ latus rectum, we have
⅛== —, and hence dy m i*V ∖ ( 2 11
«-arc AP=∕ ) 1 + Jζ ( ¾, √0 I wι 2 ) j _ yfjy ∖ +m 2 , m (y l + √ y J + m Λ 2m + 2 10g 'V m J ’ therefore whole arc PAP'— + + in l 0 g, ^i + ^yî + ^ y
Since y^≈%mx u the formula may be written arc PAP'=√4^Γ+7j + g l 0g , ∕ ⅛⅞ + √4a⅞ + yQ.
§ 48. Area of a Parabola.— Taking the equation to the parabola in the form y 2 ≈4px, we get
∕*∙z∙ 1 area of segment PAP' (fig. 21)-2ζ t λjpxdx
From these formulae we see that the area of a parabolic segment varies directly as the cube of the square root of the abscissa, and directly as the cube of the ordinate, and that it is equal to § rectangle PQQ'P ’, or ∣ triangle PTP .
A similar relation holds for the segment cut off by any chord, and thus the area of any parabolic segment can be determined in terms of any data that are sufficient to determine the segment.
§ 49. Area of a Parabolic Zone.— Let PM (fig. 22)QN = y a , AM=x 1 , AN-a⅛, and let the ordinates be inclined to the axis at an angle a.
Area of zone PQQ'P ’ —segment PAP ’- segment QAQ' -⅛Aι xβtoα . 3p
Now y 2 l — 4pa⅛ and y 2 — 4px 2 , therefore -^ 1 ~ =p, and hence on 4(α 1 - x a ) substituting for p we have area of zone
= ⅜(¾ - ⅝) (⅛≡f I) sxnα-⅜ (y 2 + y 1 y a + yξ)sinα.
C. The Ellipse.
§ 50. Circumference of an Ellipse.— The equation to the ellipse being -^ a + — 1, where a and b are the semiaxes, we have _ ⅛-, and therefore (fig. 23)
dx a 2 y /•a ( l a pe 2 x i ) ⅜ a 2 -b 2 arc of quadrant EQ=∕ j a 2- a > j where e 2 ∙=-^ 2 — .
This integral may be shown to be equal to the series
πα∕ , β 2 1 . 3e 4 1 . 3 2 . 5e β , n ∖
2∖ 2 a 2 a .4 a 2 a .4 a .6 a )'
a rapidly converging series when e is a small fraction.
By taking more and more terms of the above series we can approximate as nearly as we please to the circumference of an ellipse. For example, we have quadrant AB =ψ(ι -τH(ι -⅜) i to a first approximation; hence whole circumference ^ 7r ∣ W+(2⅛) 2 p nearl y ι
§ 51. Area of an Ellipse.— We have at once area = ∖ i^ydx^ ==— f ∖ )a?- x t dx. J o ®/o
But^ ^Ja 2 -x 2 dx is the area of the quadrant of a circle of radius a. Thus area of ellipse = 4— ∙^- (§ 44) a 4
= τrαδ.
The following proof is worth the reader’s attention. By a well-known theorem in conic sections the orthogonal projection of a circle on a given plane is an ellipse. Now, if A denote the area of any plane figure, A' the area of the projected figure, and θ the angle between the planes it can easily be shown, by dividing the two areas by planes indefinitely near to each other and perpendicular to the common section of the planes, that
A cos 0=A'.
In the case of the circle and ellipse A = πα 2 and cos0 = -; δ a hence area of ellipse = πα 2 × —— τrab.
§ 52. Area of an Ellipse in terms of a Pair of Conjugate Diameters.— Let a' and δ' denote the semiconjugate diameters, and α the angle between them, then by an elementary property of the ellipse αδ=α'δ'sinα; hence area of ellipse = τra'b' sin a.
D. The Hyperbola.
§ 53. Area of a Segment of an Hyperbola.— The equation of an hyperbola being — - 1, we have
2∕=-^-Vic 2 -α 2 ; hence (fig. 24)
⅛ f χ ∖ , area of the segment PAP' = 2 - ∕ ∖ ∕x 2 -a 2 dx CL J a,
≡-^-a⅛√^ a - αδlog.(⅛+^ΞΞ 3 )
-a⅛y 1 -αδlog β 0J-4-ll) .
§ 54. Area of a Sector of an Hyperbola.— The sector PAP'C is equal to triangle PCP' - segment PAP'
=x 1 ι∕ιI x 1 y 1 - ab}og e ^ + ^ ∣ - αδ ι°β∙(⅜ + ⅞-)∙
§ 55. Area of a Zone of an Hyperbola.— In fig. 24 the zone PP Q Q
= segment QAQ'-segment PAP'
~x i y i - αδlog^¾- + y) - * l y- i + αδlog,^ + Ç-)
-ar 2 y 9 -a5 1 y 1 -αδlog,f ^ 2 -∖ w } ιere ∖ ay l + bx 1 J x υ y 1 and x i , y 2 are the coordinates of P and Q respectively.
If the axes of coordinates be inclined at an angle α, we multiply the above results by sinα to obtain the correct areas.
§ 56. Area bounded by an Hyperbola and its Asymptotes.— The equation of an hyperbola referred to its asymptotes is of the form xy≈c 2 .
Let CM' (fig. 25)=z 1 , CM=a⅛ QM'=y 1 ,Q M = y 2 , then, if α be the angle between the asymptotes, area of QMM'Q'= ∕ * 1 y sin adx J χ ι
- cisin ^C ’ v-^ βtaαlo ≡'(5)- c,sin " ι °8∙(t) ’ c 2 , c 2 since X. ≈ — and ‰=— .
Vi Vi o α 2 + δ 2 , . 2αδ , ,, ,
Now c 2 = —.— and sin a =— 5 —j≡, and therefore
4 a 2 + b 2
(α ) area = iaδlog e ^) = ⅜aδiog i ^) .
Again, let MM'=a⅛-α⅛- p, then c 2 =.'r 1 y 1 = ¾y 2 = ^l^- , therefore
(3 ) QMM'Q'=≡⅛g.(^¼α=≡‰(^‰.
y 2 -y x fe ∖a⅛∕ 3∕ a -yι s ∖ y i J
Again, since ⅛x 1 y 1 sin a =⅛c 2 sin a ≈ ⅛x 2 y 2 sin a, we have triangle QCM = Q'CM', and hence the sector QCQ'=QMM'Q'.
The corresponding results for a rectangular hyperbola are obtained by substituting in the above formulae ⅛α 2 for c 2 and 1 for sine.
Section III.— Plane Irregular, Rectilineal, and Curvilineal Figures.
A. Irregular Rectilineal Figures.
§ 57. The area of any irregular polygon can be found by dividing it into triangles, trapeziums, &c., in the most convenient manner, and adding together all the areas. For example,
ABCDEF (fig. 26) = CKB + BKHA + AHF + FGE + EGID + DIC∙
It may sometimes happen that some of the component figures have to be subtracted instead of added; for example,
ABODE (fig. 27) = AFHE + BCG-AFGB-EDH.
§ 58. Again, the irregular rectilineal figure P 1 P 2 . . . P ? P 9 (fig. 28) can be broken up into a series of triangles and trapeziums as shown in the figure, and hence its area can be found.
§ 59. A figure made up of straight lines may be measured by cutting it up into triangles by lines drawn from some one vertex to the others. For example (fig. 29),
ABCDEF = ABC + ACD + ADE + AEF.
If the polygon be concave some of the triangles will have to be subtracted.
§ 60. Area of a Polygon in terms of the Coordinates of its Angular Points.— Let the coordinates of P, Q, R (fig, 30) be (®i, y 1 ), ( a ⅛ y 3 ), a ≡ (⅝, ⅜) respectively, and let the axes be inclined at an angle a. Draw PL, QM, and RN parallel to OY, then
LM =0M-0L =a⅛-a⅛, MN=0N -OM=α⅛-z a , and
NL≡ON-OL -3⅛-a⅛.
Now PQR=PLMQ + QMNR-PLNR.
But PLMQ = PLM + QMP = PLM + QLM
= ⅛PL∙LMsinα + iQM ’ LMsin(180°-α) (§ 9, β) = ⅜(a⅛- Xι)(yι + y 2 ) sin a .
Similarly QMNR = ⅛(a⅛ - α⅛)(y 3 ⅛ y 3 ) s in a, an d PLNR = ⅜(a⅛ - x 1 )(y 3 + y 1 ) sin d ; hence area of PQR=⅛sin α {y 1 (¾ - ⅞) + y i (a⅛ - a⅛) + y 8 (a⅛ - a ⅛)}; or in the notation of determinants [table]
When the axes are rectangular sin α=sin90° =l, and the formula for the area becomes [table]
§ 61. The area of any rectilineal figure of n sides can be found by taking any point within the figure and joining it to the n vertices of the figure, thus dividing it into n triangles the area of each of which can be obtained as in the preceding case.
We may, however, find the area of the figure directly.
For example, in fig. 31
PQRST = PPTT + TT v S'S + SS ’ R'R - RR'Q'Q - QQ'PT, and in fig. 32
PQRSTU = PP'U'U + RR'Q'Q + TT'S'S - PP'Q'Q - RR'S'S - TT'U'U.
§ 62. Length of any Curve.— If we divide the given arc into an even number of intervals and regard these as approximately circular, we can find an approximation to the length of the arc by means of Huygens’s formula, § 32. For example, if we divide ABC (fig. 33) into four parts in D, B, and E, and draw the chords AD, AB, DB, BE, BC, and EC, then arc AC = AD + DB + BE + EC + ⅜(AD + DB + BE + EC-AB-BC) approximately.
For other methods of approximation, see Rankine’s Rules and Tables.
§ 63. Area of an Irregular Curvilineal Figure.— For rough approximations the following, called the trapezoidal method, may be
Divide A 1 A n (fig. 34) into n equal parts, and through the points of division draw the ordinates, called by surveyors offsets, A 1 P 1 , A 2 P 2 , &c.
Let A 1 P 1 =s 1 , A 2 P 2 =∙s 2 , &c., AnPn=s n , and A 1 A 2 =A 2 A 3 = . . . =A n -ιA n = α.
Join P 1 P 2 , P 2 P 3 , &c., then the area of the polygon A 1 A n P n P 2 P 1 “•A·l·A·jPsPi + A 2 A3P3P 2 -f- .... + A n -lA n P n Pn-l = ⅜α(s 1 + s 2 ) + ∣α(s a +¾)+ .... + ⅛α(s n -ι + s>∣) (§ 13, α) =a{⅜(Sι+s n ) + s 2 +⅞+ . . . .+s n -ι}.
If we take n sufficiently great the difference between the area of the polygon and the curvilineal figure can be made as small as we please, in other words, the smaller we make a the more accurately will the above formula represent the area of the curvilineal figure.
The curve may either be wholly convex or wholly concave to the line A 1 A n , or partly convex and partly concave.
§ 64. Simpson's Rule.— Let A 1 A n (fig. 34) be divided into an even number of equal parts, and as before through the points of division draw the ordinates A 1 P 1 , A 2 P 2 , &c.
Let A 1 A 3 P 3 P 1 (figs. 35, 36) be a part of the figure thus divided; join P 1 P 3 , and through P 2 draw BC parallel to P 1 P 3 to meet A 1 P j in B and A 3 P 3 in C. Conceive a parabola to be drawn through P 1 P 2 P 3 having its axis parallel to the ordinates, then
A 1 P 1 P 2 P 3 A 3 =trapezium A 1 P 1 DP 8 A 3 ± parabolic segment P 1 P 2 P 3
=α(s 1 + ⅞) ± ⅜α {¾~⅛(5 1 + ⅞)} = ⅛ α (s ι + ⅜ + ⅞) ·
Now when the points P 1 ,P 2 ,P 3 are near each other the parabolic curve will coincide very nearly with the given curve; hence
A 1 PιP 2 P 3 A 3 =⅛α(sχ ÷ 4s 2 + s 3 ) very nearly.
Similarly A 3 P 3 P b A 5 - ⅞α(s 3 + 4s 4 + s 6 ), &c.; hence whole area of figure
= ⅛α{s 1 +s n +2(s 3 +s 6 + .... + s n -2) + 4(s 2 + s 4 + .... +βn-ι)}f whence the rule :—add together the two extreme ordinates, twice the sum of the intermediate odd ordinates, and four times the sum of the even ones, and multiply this result by one-third of the common distance between the ordinates; the result is the area,— accurately if the curved boundary be the arc of a parabola, in other cases approximately.
The curve may either be wholly convex or wholly concave to the line A 1 A n , or partly convex and partly concave, provided in the latter case the points of contrary flexure occur only at the odd ordinates, for otherwise the intermediate arcs could not be even approximately parabolic. When points of contrary flexure occur ordinates may be drawn at these points, and the intermediate arcs being found separately may be added to obtain the whole area.
§ 65. In the two preceding sections we investigated two formulae for approximating to the areas of curvilinear figures. We now proceed to consider the subject more generally.
Let the equation to the curve P 1 P i ,P n +ι (fig. 37) agree with the equation y = A + Bz + Cz 2 + .... + Kz n for η +1 points between P 1 and P„+i, then the area of the curvilinear figure bounded by the straight lines A 1 P 1 , A 1 A π+ ι, and A„+iP n 4-i and the curve P 1 P n ÷ι will agree very nearly with the curvilinear figure bounded by the same straight lines and the curve whose equation isy=A + Bz + Cz 2 + .... + Kz n , and the greater the number of common points the closer will be the agreement.
Let A 1 A n -∣-ι be divided into n equal parts, each equal to ħ, then A 1 A n -∣-ι = nh. Now whenz=0, y = y 1 = Aj when x—h, y=y a ≈ A + BA + C⅛ 2 + .... +KA n ; when x=2ħ, y = y 3 =A + B(2A) + C(2⅛) 2 + .... + K(2A)"; when x=ph, y=y ’,+ιA + B(pΛ) + C( pλ) 2 + .... +K(j>A) n ; when z=nA, y-y n +ι- A + B(n⅛) + C(nA) 2 + .... + K(nA) n .
From these n + 1 equations the n + 1 quantities A, B, . . . . K can be determined as functions of y 1 , y i , . . . . y n +ι » and h.
Next let A 1 A n φι be divided into m equal parts each equal to h.
Thus mh=nh and hence A=— m
Now the area of the rectangle A p A p φιP p R = A p A p +ι × A p P p .
But A p P p = y p =A + B(∕>A) + C(p⅛) 2 + .... +K(yΛ∙) n
-Λ + B*⅛ + C(⅛⅛Y + ....
m ∖ m J ∖ m J since h=—; m and A p A p+1 = ⅛=-; r r m therefore area of A p R
rih j — + BnA⅛Cn 2 Λ 2 ⅛ + . . . . + Kn n h n -ζ 7 .! . ( m m 2 m 3 m n + i )
Hence the area of the whole figure
= I. i — + Bw⅛A + Cn 2 Λ 2 ⅛ + . . + Kn"Λ"-ζr i I
χj m=∞ I m m 2 m 3 m n + 1 J
= I. n⅛i A —+ BπA⅝ + Cn 2 ⅛ 2 ⅝+ . . . . + Kn"A"⅛ ί ’ -*-Λn=∞ ( m m 2 m 3 m n + 1 ) where S n = l n +2 n + 3 n + . . + m n . Now if we take the limit of each of the terms
S o S 1 S 2 S w m * wι∙ 2 ’ m 3 ’ · · · wι n+ι »
we obtain area of curvilinear figure = n⅛ j A+ .... q— ^-n n h n ∖ . (23 n +1 )
From this general result we can deduce “Simpson’s Rule’’and also another rule called “Weddle’s Rule.”
Thus let n= 2; that is, assume that the curve under consideration has three points in common with the curve whose equation is y=A + Bz+Cz 2 , i.e., with a parabola, then
?/i = A, y 2 = A + BA + CA 2 , y 3 = A + 2 BA + 4CA 2 .
Now the area is approximately
= 2A{A+⅛B2A + ⅜C2 2 A 2 }
= ⅛A{6A + 6BA + 8CA 2 }
= ⅜Λ∙ {y 1 + 4y 3 ÷ 3∕ s } , Simpson’s Rule.
If we now put n = 6, we have area of curvilinear figure
= 6A { A + ⅜B6A + ⅛C6 2 Λ 2 + ∣D6 3 A 3 + ⅜E6 4 Λ< + ⅛F6 5 A 5 + ∣G6 β A β }.
Now y 1 = A,
⅜=A + BA + CA 2 + .... + GA β , y 7 ≡A + B(6A) + C(6A) 2 + .... G(6A) β .
From this system of equations we can determine A, B, C, . . . G, and substituting the values so obtained in the above expression we obtain the following remarkable formula for the approximate area: area≈⅛Λ{(y 1 + y 3 + 2 ∕ 5 + y 7 ) + y 4 + 5(y i + y 4 + y β )j .
This formula, called Weddle’s Rule, gives the closest approximation to the curvilinear area that can be obtained *
by any simple rule.
We are now in a position to find the approximate area of any irregular plane figure. For the given figure can be divided into plane rectilinear and curvilinear figures, the areas of which can be separately determined by the rules already given. For example, APQRS (fig· 38)
- ABC + APD + BRC - DQB - ASC.
PART II. SOLIDS.
Section I. Solids contained by Planes.
A. Prisms, Pyramids, and Prismatoids.
§ 66. Volume of a Right Prism. —First let the prism be a rectangular parallelepiped (fig. 39), and let the side AB contain a units of length, BC b units of length, and CD c units of length. If we divide AB into a equal parts, BC into b equal parts, and CD into c equal parts, and if, through the points of division we draw planes parallel to the sides of the parallelepiped, these planes will divide it into a series of parallelepipeds, whose edges are each equal to the unit of length. Each horizontal layer contains ab of these cubes, and since there are c layers the whole number of cubes will be abc. But each of these is the unit of volume, and therefore volume of ABCD =abc=ab × c=area of base ABC × altitude c.
In the above demonstration we have assumed the edges to be commensurable, but from § 2 it follows that the proof will hold also when the edges are incommensurable. If the parallelepiped be cut by a plane BGE it will be divided into two equal triangular right prisms, and hence volume of right triangular prism = ⅜α⅛ × c=area of its base × altitude. Since every prism can be divided into triangular prisms as in fig. 40, we have at once volume of right prism A'ABCDE=A'ABC + A'ACD + A'ADE = ABC X BB' + ACD × CC ’ + ADE × DD'
— (ABC + ACD + ADE) × altitude
(since BB' = CC' = DD'=altitude)
= area of base ABCDE × altitude.
§ 67. Volume of an Oblique Prism.— Draw the right section A"B"C"D"E" (fig. 41), and let A' denote its area and A the area of the base A'B'C'D'E'. Let I denote the length of the prism, h its altitude, and α the angle between the planes A'B'C , DΕ' and A"B"C"D"E".
Conceive the part above the right section placed at the other extremity of the prism. Then we have a right prism, whose volume = A' × I (§ 66); but A'= A cos a, since A' is the projection of A (§ 51), and; hence cosα
volume = A'×Z=Acosa×-^-=A×A: cos a or the volume of any prism is equal to the area of its base multiplied by its altitude.
§ 68. Surface of a Prism. —Since the lines A"B", B"C'', &c. (fig. 41), which make up the perimeter of the right section are all in one plane perpendicular to the parallel edges A ’ A" ’, B ’ B , ", &c., they are perpendicular to these edges and are therefore the altitudes of the parallelograms A ’ B'B" ’ A , B'C'C"'B" ’, &c., respectively. The lateral surface of the prism is equal to the sum of these parallelograms, and therefore
- A ’ A'" × A"B" + B'B' × B"C" +
— A'A ’ "(A"B" + B"C" + ), since A'A" ’-B'B"'-fcc.; or the lateral surface of any prism is equal to the perimeter oι its right section multiplied by the length of the prism.
If the prism be right, that is, if the faces be perpendicular to the base, then its lateral surface is equal to the perimeter of its base multiplied by its length.
The whole surface of any prism is obtained by adding to the lateral surface the areas of its bases.
§ 69. If the prism be regular, that is, if the bases be regular polygons, then area of base —α 2 ×cot ?^-(§ 18, 7), where n is the number 4 n of sides each of length a, and therefore volume —α 2 × — cotl^- × h,
4 n where h is the altitude of the prism.
Again, if the prism be right and regular, then its lateral surface — nah + 2α 2 × —cot^- .
4 n
§70. Volume of a Pyramid.— Let VABC (fig. 42) be for simplicity a triangular pyramid. Divide V A into 11 equal portions, and through the points of section draw planes parallel to the base ABC, and through BC and through the intersections of these planes with VBC draw planes parallel to VA. Let h denote the altitude of the pyramid, then the distance of the base of the r th prism from the vertex V
h =r×— , n and, if A denote the area of ABC, we have base of r th prism r 2 h 2 1 _ r a A n 2 A 2 ri 2 ’
since, by a well-known theorem in solid geometry, the areas of sections of a pyramid made by planes parallel to the base are proportional to the squares of their altitudes.
Thus we have base of r th prism = 2ζ-A, and therefore τι a its volume = -ζ-A × — (§ 67) n i n hλ 2 — — 5 - × r l . w 3
Therefore volume of whole pyramid
-*λl,√ ’ ÷*÷ · · · ·
-⅞AL,,,. ιl <" +1 6 ^÷ 1 >=⅛× ii or the volume of any pyramid is equal to one-third of the area of its base multiplied by its height
From this we see that pyramids on equal bases are to one another as their altitudes.
If the pyramid be regular, that is, if its base be a regular polygon the perpendicular through whose centre passes through the vertex, its volume — ⅛ × α 2 × 2Lcot^^- × h.
4 n
§ 71. Surface of a Regular Pyramid.— The lateral surface of the regular pyramid VABCDEF (fig. 43) is equal to the sum of the areas of the n congruent triangles which make up the lateral surface of the pyramid.
Now area of triangle VAB = ⅜AB × VG; hence whole lateral surface— ⅜nAB - VG = ⅜nαZ, where I is the slant height and a the length of the side of the base.
Again, if VO = ⅛ = altitude of pyramid, we have
Z-VG-√VO 2 + OG 3 - Jk i + ∙j∙ cot 2 — , therefore whole surface — base + lateral surface 2 n 180° 1 . ∕, 2 1 α 2 ,,180“
—a 2 × —cot h ∖ na a , ∕ ½ 2 + - 7 - cot 2
4 n 8 V 4 n naf a . 180° ∕ , 2 , d i , 2 180° ∖
= - s -( —cot + a. / ⅛ 2 + -rC0t 2 .
2 ∖ 2 n ∖ 4 nJ
§ 72. The Prismatoid.— If we have a polyhedron whose bases are two polygons in parallel planes, the number of sides in each being the same or different, and if we so join the vertices of these bases that each line in order forms a triangle with the preceding line and one side of either base, the figure so formed is called a “prismatoid,” and holds in stereometry a position similar to that of the trapezium in planimetry. To make the investigation of the volume of the prismatoid as simple as possible, we take the case where the lower base is a polygon of four and the upper one 01 three sides.
Let ABCDEFG (fig. 44) be the prismatoid, of which ABC or A 1 is the upper and DEFG or A 3 the lower base, and let HLM be the section equidistant from the bases. Take any point P in this section and join it to the corners of the prismatoid. We thus divide the polyhedron into two pyramids PABC and PDEFG, and a series of polyhedra of which CPDE may be taken as a specimen.
Let h be the altitude of the prismatoid, then ⅜½ is the altitude of each of the pyramids PABC, PDEFG, and hence volume of PABC — ⅜ΛA 1 , and volume of PDEFG —⅜ΛA 3 .
Again join PH, PL, and LD, then volume of CPDE = 2 volume of CPDL, since DE = 2HL, and volume of CPDL —2 volume of CPHL, hence volume of CPDE —4 volume of CPHL.
Now volume of CPHL —⅜Λ × area of HLP, and therefore volume of CPDE — ⅛ h × area of HLP.
Similarly the volume of every such polyhedron is ⅜Λ × the area of its own portion of the middle section. Hence if A 2 denote the area of the middle section we have volume of prismatoid — ⅜ΛAι + ⅜AA i , + ⅛λA 2
— ⅞Λ(A 1 + 4A 2 + A 3 ).
§ 73. Volume of the Frustum of a Pyramid.— Let A'A"'B'B"'C'C"' (fig. 45) be a frustum of the pyramid VA'B'C', and let A 1 and A 3 denote the areas of the ends A'B'C', A"'B"'C"' respectively. Let VP— x≈ altitude of pyramid VA"'B"'C"', and let PQ — h — altitude of frustum.
xτ ∕ a: ∖ 2 A 3 Έ A√A s
Now —- 7 I —- r i , whence a:——77— δ —77-.
∖ x+hj A 1 ’ √A 1 -√A 3 Again frustum = VA'B'C' - VA'"B ’ "C"
= i∣ A 1 (as+Λ)-A 3 ae j∙
= έ { λl (√A 1 ^ √A 8 ) ~ a3 (√A 1 ^ √A 3 ) }
= ⅛Λ(A 1 +VA 1 A 3 + A 3 ); a formula which applies to the frusta of all pyramids regular and irregular.
The above result may be otherwise expressed. For, let A ’ B'=α 1 and A" ’ B"'=α 3 , then, if A"B"C" be a section equidistant from the ends of the frustum, A"B"=α 2 =⅛(α 1 + α 3 ).
Now R i ≈pa i l and A 3 =2?a| (see § 70); hence A 2 = area of A"B"C" =pa i i =p(^ , which gives
4A 2 =pa ∖ + 2pα 1 α 3 +pa 2 s =A 1 + 2√A 1 A 3 + A 3 ; therefore volume of frustum
= ⅜A(2A 1 + 2√A√Γ 3 + 2A 3 ) - ⅜A(A 1 + 4A 2 + A 3 ); or the volume of the frustum of a pyramid is obtained by adding the areas of the ends to four times the area of the middle section, and multiplying the sum by one-sixth of the altitude.
The above result can be obtained at once from § 72, since A'B'C'A'"B"'C"' is a prismatoid with similar bases.
§ 74. Surface of the Frustum of a Regular Pyramid.— In fig. 45 let the perimeter of A 1 =j7 1 , that of A 2 =∕7 2 , and that of A 3 =‰ and let VD'=Z 1 , VD"' = Z 3 , and therefore D'D'"= VD' - VD" ’- l 1 - l 3 = I. The lateral surface of the frustum is equal to the difference between the lateral surfaces of the pyramids VA'B'C' and VA"'B"'C"',
-i⅛-fo⅛⅛∙
But, since φ- = -= -—= ^ we have l 1 ≈~ 1 ^ ’ and Z g = , Z 3 a 3 m 3 p 3 ’ P 1 ~P 3 3 Pι~Ps therefore lateral surface of frustum
=<≡D=≈m→-
or the lateral surface of the frustum of a regular pyramid is equal to the product of the slant height and the perimeter of the section equidistant from the ends.
Otherwise.—The top and base being regular polygons, the inclined faces are congruent trapeziums. Let I be the height of each trapezium, ana let there be n of them, then area of each face = 4-f—+^ s Λ , 2 ∖ n nJ and therefore the area of lateral surface = -i- (p 1 +p 3 ) = lp 3 .
§ 75. If h the altitude of the frustum be given, we deduce the slant height and then proceed as before. Thus let VP=λ 3 , VQ=Λ 1 , and using the same notation as in §§ 72, 73, and 74 we have r ⅛ ⅛-, which gives⅞ 1 , α ι<⅜-¾)-¾I. .
Λ 1 - Λ 3 α 1 - α 3 a ι~ a 3 a ι ~ a 3
Again ^=^ + ⅛αJcot 2 ^, and Z 1 ; whence I is known since l 1 is known in terms of h.
When the pyramid is irregular the lateral planes are non-congruent trapeziums, the areas of which can be found separately by § 12, and hence the whole surface.
§ 76. Volume of the Frustum of a Triangular Prism.— Let A denote the area of ABC (fig. 46), and let h 1 , h 3 , h 3 be the altitudes of A', B', C' respectively with reference to the plane ABC. Divide the frustum into three pyramids B'A ’ AC, B'ABC, and B'A ’ CC' by the planes B'AC and B'A ’ C. These three pyramids are respectively equal to BA'AC, B'ABC, and ABCC'; hence volume of frustum — ⅜7t 1 A + ⅜⅛ 2 A + ⅛⅛ 3 A
=⅜A(Λ 1 + Λ 2 + Λ 3 ) .
§ 77. If the prism be right or oblique, the volume of a frustum is equal to one-third of the area of its right section multiplied by the sum of the parallel edges. For divide the frustum AA'B'C ’ (fig. 47) into two frusta by a plane A"B"C" of area A at right angles to the edges, then
AA'B'C'=AA"B"C" + A"A'B'C ’
= ⅜A(AA" + BB" + CO") + ⅜A (A ’ A" + B'B" + C'C")
=⅛ A(AA" + BB" + CC" + A'A" + B'B" + C'C")
= ⅜A(AA' + BB' + CC').
Again, since every prism can be divided into triangular prisms, we can find by repeated applications of the above proposition the volume of the frustum of any prism whatever. For example, if the base of the frustum of a right prism AA'B'CΙ)' (fig. 48) be a rectangle 12 feet by 6 feet, and the parallel edges in order 6, 4, 10, and 12 feet, then
A = area of base = 12×6 = 72 square feet. Frustum = ABC A'B'C' + ADCA'D'C'
= J × ⅛A(AA' + BB ’ +CC ’) + ⅜ × ⅜A(AA' + CC'+DD')
- ⅛ A (2 AA' + 2CC' + BB'+DD') = 576.
§ 78. Volume of a Wedge.— The wedge (fig. 49) being merely the frustum of a triangular prism, we have at once volume = ⅜A(FE
+ AD + BC), where A is the area of its right section; otherwise, the wedge may be considered a prismatoid whose upper base is a straight line, and hence its volume = ⅜Λ(4A 2 + A 3 ), since A 1 =0.
B. Regular Polyhedra.
§ 79. The regular polyhedra are five in number, namely, the tetrahedron, cube, octahedron, dodecahedron, and icosahedron, whose solid angles are formed respectively by three equilateral triangles, three squares, four equilateral triangles, three pentagons, and five equilateral triangles.
Since a regular polyhedron admits of having a sphere inscribed within it and described about it, it can easily be shown that the volume of the polyhedron cos— cot 2 —
„ nl m n x ≡ x j . cm ∕2L + '.) co ,(Z.-«.)}· ’
( ∖ m nJ ∖ m nJ) and from § 18, y, it follows that the surface of the polyhedron „ In , ιr
= α 2 ×- r cot—, 4 n where I =the number of faces,
77i=the number of faces in each solid angle, 7i = the number of edges in each face, and α=the length of each side.
The following table contains the surfaces and volumes for the five regular polyhedra whose edge is 1. [table]
The surface and volume of a regular polyhedron whose edge is a is obtained by multiplying the surface and volume of a similar polyhedron whose edge is 1 by α 2 and a 3 respectively.
Section II. Solids contained by Surfaces which are not all Planes.
A. The Cylinder.
§ 80. Volume of a Cylinder (fig. 50).—Inscribe in the cylinder a polygonal prism of which the number of sides may be increased indefinitely. Then in the limit the base of the prism becomes the base of the cylinder, and the volume of the prism the volume of the cylinder. Now by § 67 we have volume of prism — area of base × altitude; hence volume of cylinder=area of base × altitude.
§ 81. Surface of a Right Cylinder.— As above, in the limit the base of the prism becomes the base of the cylinder, and the surface of the prism the surface of the cylinder. Now the lateral surface of prism
— perimeter of right section × length
—perimeter of base × length, in the case of a right prism (§ 68); hence lateral surface of right cylinder=circumference of base X length.
§ 82. Volume of a Cone (fig. 51).—Inscribe within the cone a pyramid of which the number of sides may be indefinitely increased, then in the limit the base of the pyramid becomes the base of the oone and the volume of the pyramid the volume of the cone. By § 70 volume of pyramid = ⅜ base × altitude, and hence volume of cone = ⅞ base × altitude.
§ 83. Volume of the Frustum of a Cone.— From § 73 we find that the volume of the frustum of a pyramid
= ⅞A (A 1 + √A 1 A 3 + A 3 ); hence, since in the limit the frustum of the pyramid becomes the frustum of the cone, we have volume of conical frustum — ⅛h ( A 1 + √A 1 A 3 + A 3 ), where A 1 and A 3 are the areas of the terminating planes of the frustum.
Let the terminating planes be circles of radii r 1 and r 3 , then volume of frustum
= ⅜⅛ (wrJ + ιrr 1 r 3 + nr*) ≈ ⅜π7i (rj + r 1 r 3 + r i 3 ).
Again, by the same section we have volume of frustum of pyramid=⅜½(Aι + 4A 2 + A 3 ), therefore volume of conical frustum=⅜πΛ (r J + 4>r 3 + r%), where r 3 is the radius of the circular section parallel to the terminating planes and equidistant from them.
§ 84. Surface of a Right Cone.— The lateral surface of a regular pyramid is by § 71
= ⅛ perimeter of base × slant height; hence, since in the limit the surface of the pyramid becomes the surface of the cone, the lateral surface of a right cone is equal to half the circumference of its base multiplied by the slant height.
Thus the lateral surface of a right cone of slant height I and the radius of whose base is r is equal to
⅜ X 2πr × Z = πrZ, and whole surface = lateral surface + area of base = τrrl + τrΓ 2
= πr(Z + r).
Again, if h, the altitude of the cone, be given, we have l≈^∕hf^, and therefore whole surface = πr(√Λ 2 + r 2 +r).
§ 85. Surface of the Frustum of a Right Cone.— The lateral surface of the frustum of a regular pyramid is equal to the product of the slant height and the perimeter of its middle section (§ 74); hence in the limit we find that the lateral surface of the frustum of a right cone is equal to the product of its slant height and the circumference of tne section equidistant from its parallel faces.
Let r 1 and r 3 denote the radii of the ends of the frustum, and I the length of the slant height, then r a"⅛(r ι + r s) i = radius of middle section, and therefore lateral surface = 2πr 2 × I =2ιr × ⅜(r 1 + r 3 ) × I - π Z(r 1 + r 3 ), and whole surface— ιrτ* + πZ(r 1 + r 8 ) + τr 3 .
If h, the altitude of the frustum, be given, we have Z-√Λ s + (r 1 -r 8 ) ’.
C. The Sphere.
§ 86. Surface of a Spherical Zone.— Let AB (fig. 52) be a small arc of the sphere, and let AA ’, BB' be perpendicular to the axis XX', to find the surface of the zone generated by the arc AB. Join AB, and draw OP perpendicular to AB, BD parallel to XX', and PP' parallel to AA' or BB'. The chord AB generates the frustum of a cone, whose lateral surface = 2πPP'×AB.
But, since the triangles ABD and OPP' are similar,
AB = OP BD PP' ’
therefore area of conical frustum — 2τr.OP. BD = 2π. OP. A'B'. Similarly the area of the frustum generated by BC = 2τr. OQ. B'C'.
But in the limit when the chords AB, BC, &c., are indefinitely diminished, the perpendiculars OP, OQ, &c., become each = r, and hence by summing all the areas we get in the limit.
area of zone = 2πr × (projection of arc on axis of revolution). Hence the convex surface of a segment of a sphere is equal to the circumference of a great circle multiplied by tne height of the segment or zone.
§ 87. Surface of a Sphere.— The whole sphere being a zone whose height is 2r, we obtain at once surface of sphere = 2ιrr × 2r=4πr 2 ; or the surface of a sphere is equal to four great circles.
The total surface of the cylinder circumscribing the sphere of radius r is 6ττr 2 , hence the surface of the sphere = ∣ surface of circumscribing cylinder.
§ 88. Surface of a Lune, a Spherical Triangle, and a Spherical Polygon.— It is shown in spherical trigonometry that
(a) the area of a lune included between two great circles of a sphere of radius r, and whose inclination is θ radians, is
20r 2 j
(3) the area of a spherical triangle whose angles are A, B, C is (A + B + Cπ)r 2 j
(γ) the area of a spherical polygon of r sides is
{P - (r- 2)ιr}r 2 , where P is the sum of its angles.
§ 89. Measurement of Solid Angles.— A convenient unit for the measurement of plane angles is the “radian.” If we assume that each unit of surface of a sphere subtends the same solid angle at the centre, we can deduce a very convenient unit for the measurement of solid angles. This unit, which has received the name “steradian,” we define to be the solid angle subtended at the centre of a sphere by a portion of the surface whose area is r 2 .
§ 90. Number of Steradians in an Angle.— Let A be the angle at the centre of a sphere, and let S be the portion of the surface of the sphere which it intercepts, then number of steradians in A S
1 ""i 3 ^ ’
For example, if A be a plane solid angle, S = a hemisphere = 2πr 2 ; hence the number of steradians in a plane solid angle = 2π, r a ι a and therefore the number of steradians in the solid angle at apoint = 4τr. This solid angle is sometimes called a steregon.
Hence, if we can find the surface subtended by any solid angle, we can always find its magnitude in terms of the unit solid angle.
§ 91. Volume of a Sphere.— Let ABC
(fig. 53) be the quadrant of a circle, draw DB and DC tangents to it, then, if AD be joined and the whole figure be conceived as rotating round AB, ABD, ABC, and ABDC will generate a cone, a hemisphere, and a cylinder respectively.
Now draw two parallel planes EFGH and E'F'G'H' very near each other and perpendicular to AB, and draw FF' and GG' parallel to AB, then, by § 80, volume generated by EHH'E'=ιrEH 2 x EE',
„ „ EGG'E' =ιrEG 2 ×EE',
„ ,, EFF'E' =τrEF 2 ×EE'.
Thus volume generated by EFF'E'+ volume generated by EGG'E' =π(EF 2 + EG 2 ) X EE' = π(EA 2 + EG 2 ) × EE'
= π(AG 2 ) × EE' = ιrEH 2 × EE'
= volume generated by EHH'E'.
Therefore in the limit, when the number of slices is indefinitely increased, and their thickness indefinitely diminished, we have volume of cone generated by AF + volume of spherical zone generated by CG = volume of cylinder generated by CH.
Let r=radius of sphere, Λ = AE=height of zone ACGE, then volume of cone = ⅛ιr⅛ 2 × Λ= ⅛τΛ 3 , and volume of cylinder=πr 2 × h, therefore volume of spherical zone=πr 2 A- ⅛%A 3
= ⅛ιτ⅛(3r 2 — A 2 ).
The height of a hemisphere is r, therefore volume of hemisphere=⅜ιrr (3r 2 - r 2 ) = ∣ πτ 3 , and volume of whole sphere=⅜ιrr 3 , a result readily obtainable by the infinitesimal calculus, or by inscribing within the sphere a series of triangular pyramids whose vertices all meet at the centre of the sphere, and the angles of whose bases all rest on the surface. In the limit the altitude of each pyramid becomes the radius of the sphere, and the sum of the bases of the pyramids the surface of the sphere; hence volume = ⅛S × r=⅛ × 4ιrr 2 × r= ⅜πr 3 .
The volume of the circumscribing cylinder=πr 2 × 2r=2πr 3 , therefore volume of sphere = ∣ volume of circumscribing cylinder.
§ 92. Let S denote the surface of a sphere and V its volume, then from §§ 87 and 91 we have
∕ ∖ 3/3 √<∕vΓ
<∙> ’ ∙-2√√-√S xvvi
(3) S = Λ v ∕ ιr (6V)≡j
ω v - 5 ⅛w i formulae which give the radius in terms of the surface or volume, the surface in terms of the volume, and the volume in terms of the surface.
§ 93. Volume of a Spherical Shell,.— Let r and r 1 denote the radii of the two spheres, then volume of shell = V = ⅞ιrrj - ⅜πr 3
= ⅜π(r≡-r 3 )
=⅜"∙frι - r){r ∖ + r 1 r + r 2 ).
Now let r 1 -r=⅛, then v - i "≡ i (ι+⅞+⅛)∙
If h be small compared with r 1 , then r ∣ r 1 is very nearly equal to 1, and we have approximately
V = ⅜πτ∙JΛ(l +1 +1) - 4πr ↑h.
Again, if h be nearly equal to r 1 , r is very small, and r∕r 1 is also very small, so that we have approximately V-⅛πr 2 l h.
§ 94. Volume of a Spherical Segment.— Let CRC' (fig. 54) be a section of a spherical segment whose altitude RQ is p, then, if OQ = Λ, volume of segment CRC'=volume of hemisphere - volume of zone AA'C'C
= Sτr 3 -⅜πΛ{3r 2 -Λ 2 }, §91.
= § πr 3 - ⅜ιr(r -p) {3r 2 - (r -p) 2 }
= ⅜πp 2 (3r -p) .
If we putjo = 2r, we obtain as before volume of sphere = ⅜πr 3 .
Again if CQ = α 1 we have
CQ 2 =αj = RQ.R'Q -ρ(&-ρ}, whence r - a > + ∙P 8 2p therefore volume of segment = ⅜ιrp (3aJ +p 2 ).
§ 95. Volume of a Spherical Frustum.— When one of the terminating planes passes through the centre we have already found that the volume
= ⅜π⅛(r 2 -A 2 ), where h is its altitude.
Now suppose that neither of the terminating planes passes through the centre; for example, to find the volume of the frustum BB'C'C.
Let RQ=p and RP=g r , then
BB'C'C = segment RBB'-segment RCC' = iπq(3a 2 + q 2 )- ⅜πp(3αj +ji 2 ), where α 1 and α 2 are the radii of the ends CC' and BB'.
Let q-p=h = height of frustum, and, since, from the geometry of the figure,
AeÎ+.P 2 α* + g 2 2j , P 1 we have volume= ⅜π⅛{3(αj + αj)+ ⅛ 2 }, a result which may also be obtained by considering BB'C'C as the difference of the two zones AA'C'C and AA'B'B.
D. Spheroid.
§ 96. Surface of a Prolate Spheroid.— The prolate spheroid is the solid generated by the revolution of an ellipse about its major axis. If S be the surface generated by an arc of the curve, then S>=2πjy + dx, taken between proper limits.
In the case before us
S = 2πδ 2 + 2π- sin ^ 1 e, e where e is the eccentricity (Infinitesimal Calculus, art. 179).
§ 97. Surface of an Oblate Spheroid.— The oblate spheroid is the solid generated by the revolution of an ellipse about its minor axis (fig· 55)∙
Here surface=2πα 2 + π--log e ii^ (Infinitesimal Calculus, e 1 — e art. 179).
§ 98. Volume of a Spheroid.— We have volume of prolate spheroid
-√> (ι -⅛>-w∙
Similarly volume of oblate spheroid=⅜wα 2 6.
Thus,
. volume of prolate spheroid ⅜wα⅞ 2 b . a volume of oblate spheroid ⅜ιrα⅛ a ’
. sphere described on major axis _ ⅜πα 3 _ A 2 .
'' fi prolate spheroid ⅜ιrαft 2 ⅛ 2 ’
. . sphere described on minor axis _ ⅜πδ 3 _ 5 2 w oblate spheroid ~ ⅛rα 2 5 α 2
§99. Volume of a Segment of a Spheroid. (a) The prolate spheroid.— This segment is generated by the revolution of AMP (fig. 23, p. 20) about AM, and hence its volume = τr∕* y 2 dx=f (2αx - z 2 )d x = 4r × (3α - h),
√o a Jg ύ a where A is the origin and AM=Λ.
(β) The oblate spheroid.— The segment in this case is generated by the revolution of BMP (fig. 55) about BC, and hence its volume = π∕* y 2 dx=∙κ^ f h {2bx-aι i )dx≈^× t ^-(fib-h), o“Jq o tr where B is the origin and BM = h.
§ 100. Volume of the Frustum of a Spheroid when one of the Terminating Planes passes through the Centre.
(a) The prolate spheroid.— The frustum in this case is generated by the revolution of BCMP about CM (fig. 23).
Now volume generated by BCMP
=volume generated by BCA
- volume generated by PMA
= ⅜παδ 2 --J×^(3α-A)
=2L×^(3α 2 -* 2 ), where k =CM - height of frustum ≈a-h.
(3) The oblate spheroid.— We can show in a similar manner that the volume generated in this case
= -J×^(3δ 2 -* 2 ).
The above formulae may be put into another form. Thus, in the case of the prolate spheroid, since the point P lies on the ellipse 5⅛ 2 + α 2 y 2 = α 2 5 2 , we have
δ 2 ⅛ 2 + α 2 ⅛J=α 2 ⅛ 2 , where 5 1 = PM, which gives whence, by substitution, the volume of prolate frustum
- i%⅛(2δ 2 + δj).
Similarly we can show that the volume of the oblate frustum
— ⅜τ⅛(2α 2 + α}), where α 1 — PM .
These formulae play an important part in the gauging of casks.
E. Paraboloid.
§ 101. Surface, of a Paraboloid.— Let the equation to the parabola be y s =4ax, and let the coordinates of P (tig. 21, p. 19) be x 1 , y,, then the surface of the paraboloid generated by the revolution of AM about AP
-∣π√α {(x 1 + α)^ - a ∣ } .
§ 102. Volume of a Paraboloid.— With the same notation we have
∕"j∣ ∕*X ∖ volume = π∕ y-dx= l∙κa∕ xdx — ⅛π × 4ax 1 × x 1 — ⅛ τry↑ × x 1 ;
-∕0 √o or the volume of a paraboloid generated by the revolution of a part of a parabola between the vertex and any point is equal to half the volume of the circumscribing cylinder.
§ 103. If the coordinates of Q be x 2 , y v then the volume of the frustum PP'Q'Q
- i» { y'ι x i - y ’ , x ι } - 2ιrα(xJ - x↑ ) - ⅛ιr(y a + y ∖ )h , where Λ = MN; hence the volume of the frustum of a paraboloid is equal to half the sum of the areas of its ends multiplied by its height.
F. Ellipsoid.
§ 104. Volume of an Ellipsoid.— The equation to the ellipsoid being α 2 + ⅛a T c 2 »
the equation to the elliptic section at the distance z from the origin is
α 2 + b* c a
Now if we draw an indefinite number of parallel planes perpendicular to the axis of z, each slice will be an infinitely thin cylindrical plate, and accordingly the whole volume of the ellipsoid =J'kdz, where A is the area of the elliptic section.
But A— 1 rαδ(l--J), § 51, therefore volume — τabj' ^dz — ⅛ιrdbc .
The sphere being an ellipsoid whose axes are all equal, we obtain as before volume of sphere = ∣ιrα 3 — ⅜πr 3 .
G. Hyperboloid.
§ 105. Volume of an Hyperboloid.— The hyperboloid is generated by the revolution of the hyperbolic segment ANP about AN (fig. 24, p. 20). If the coordinates of P be x 1 , y 1 , then
∕*∙ r ι δ 2 ∕*
volume of hyperboloid —1√ y 2 dx — τ ~ff∕ a * 1 (χ2 ~ α s )c⅛
δ 2 ( x? „ 2α 8 ) wδW,- u . jn
-v> I « - a% · + ~ J -^-< 3o+a >' where Λ=AN=x 1 - a.
Again, since x 1 , y 1 is on the curve, we have
α 2 yJ-δ 2 (α + Λ) a - - a 2 δ 2 , which gives; whence a i 2aΛ + hr volume of hyperboloid — × ·
3 2λ 4" Λ
H. Solids to which the “Prismoidal Formula” applies.
§ 106. It was shown in § 72 that the volume of any polyhedron bounded by two parallel planes and by plane rectilinear figures
“i^( A 1 + 4A 2 + A 3 ), where A 1 , A 3 , and A 2 denote respectively the areas of the two ends and of the middle section.
We now proceed to show that the same formula determines the volumes of all solids bounded by two parallel planes, provided tho area of any section parallel to these planes can be expressed as a rational integral algebraic function of the third degree in x, where x is the distance of the section from either plane.
Let ≠(x)-A + Bx + Cx 2 + Dx 3 + .... + Kx n denote the area of the section in question.
Now the solid between the sections ≠(0) and <∕>(4) is equal to the solid between the sections φ(0) and φ(2) plus the solid between the sections ≠(2) and ≠(4). Hence if the prismoidal formula is to hold in this case, we have
⅛Λ{≠(0)+ 4≠<2) + ≠(4)}
-⅛⅛{ φ (0)+ 4≠(1) + ≠(2)} +Λ⅛{≠(2) + 4≠(3) + φ(4)} , where h is the distance between the sections ≠(0) and φ(4). Hence we have
≠(0) - 0(1) + β≠(2) - 0(3) + ≠(4) - 0.
Now φ(0)-A
-4≠(1)--4A-4B-4C-4D-4E- .... -4K
+ 6≠(2)-6A + 12B + 24C + 48D + 96E + .... + 6∙2"K -4≠(3)= - 4A - 12B - 36C - 108D - 324E - .... -4 3"K + φ(4)-A + 4B + 16C + 64D + 256E + .... +4 1 *K.
Therefore 0-0 + 0 + 0 + 0 + 24E + PF+ .... +TK.
Hence E = F= . . . . K = O, and therefore φ(x} must be a function of the third degree in order that the prismoidal formula may apply.
§ 107. If we take φ{x) — A + Bx + Cx 2 + Dx 3 , there will be as many possible varieties as there are combinations of four things, one, two, three, and four together, i,e, t 2 4 -l-15 varieties. Corresponding to each of these there will be at least one solid the area of a section of which at a distance x from one of the parallel planes is φ{x) — A + Bx + Cx 2 + Dx 3 , and at least one solid of revolution generated by the curve whose equation is of the form try 2 — φ(x) — A + Bx + Cx 2 + Dx 3 .
As space prevents us discussing all the cases that may arise, we content ourselves by giving three examples as illustrations.
(a) Volume of an ellipsoid.— Here ≠(x)-Bx + Cx 2 .
Let 2a, 2δ, and 2c be the axes of which 2α is the greatest, then Λ = 2α, A 1 = 0 1 , A s = 0, and A a = πδc; therefore volume -= ⅜Λ( A 1 + 4 A a + A s ) — ∣α(4ιrδc) — ⅜ταδc, which agrees with the result in § 104.
(3) Volume of a sphere. — Here ∙κy 2 — ≠(x) — Bx + Cx 2 .
Let r be the radius of the sphere, then h = 2r, A 1 =0, A 3 = 0, and A 2 - ιrr a , hence, as before (§ 91),
— 2z*
volume of sphere = ⅜Λ( A 1 + 4A 2 + A 3 ) ——(4πr 2 ) — ⅜ιrr 3 .
(γ) Volume of a right circular cone. — Here πy 2 =≠(x) = Cx 2 .
Let r = radius of base and h the altitude, then A 1 -0, A 3 = πr 2 , and A 2 - u∙(⅜r) a ; hence volume of cone = ⅜A{A 1 + 4A 2 +A 3 } = ⅜⅛{ιrr 2 + nτ 2 } = ⅛Λιrr 2
In a similar manner we can determine the volumes of a cylinder, a prolate spheroid, an oblate spheroid, &c.
§ 108. In general, if in any solid we have
≠(x)-A + Bx + Cx 2 + Dx 3 , where A, B, C, and D are known constants, then, if h be the length of the solid,
A 1 -≠(0)-A,
A 2 - ≠(⅛Λ) - A + B(⅜A) + C(⅛Λ) 2 + D(⅜A) 3 , A 8 ≈≠(Λ) = A + BA + CA 2 + DA 3 , and therefore volume of solid—⅜Λ(k 1 + 4A s + A s )
-AA + 4BΛ a +iCΛ 3 + ∣DΛ 4 .
I. Solids of Revolution in General.
§109. Volume of any Solid of Revolution.— Let P 1 P 2 . . . . P m (fig. 34) be the generating curve, and A 1 . . . . A„ the axis of revolution. Divide the curve into portions in the points P 2 , P 3 , &c., and draw the chords and tangents of the small arcs P 1 P 2 , P 2 P 3 , &c., then it is evident that the solid generated by the curve is greater than the sum of the conical frusta traced out by the chords and less than the sum of the conical frusta traced out by the tangents. Hence, by increasing the number of chords, namely, by increasing the points of division of the curve, we can make the difference between these sums as small as we please, and therefore by this method we can approximate as closely as we please to the volume of the solid generated.
Assuming that the points P 1 , P 2 , P 3 are so near each other that the solid generated differs little from the frustum of a cone, and using the same notation as in § 63, we have volume generated by
P 1 P 2 P 3 - ^A 1 A s (sJ + 4s≈ + sj) = ⅜tγ2A 1 A 2 (sJ + ⅛ + ⅛)
— ⅜ιrα(sj + 4s*+βj); similarly the volume generated by
P3P4P5 - ⅛*Φ5 + + «Ï); whence the volume generated by the whole curve P 1 P 2 . . . . P n = ⅜ιrα{βi + ⅛ + 2(sJ + si+ .... + ⅛- 3 ) + 4(s* + sJ + - · · +4-1)}; or (since irs? = —— , τrs 2 = ·, &c.)
4τr 4π
-~{cJ + ⅛ + 2(⅛+cι + ' · · + c "-2) + 4 (c * + c 4+ · · · · + c "-ι)b a formula more convenient in practice, as it is sometimes more easj to measure equidistant circumferences than equidistant radii.
J. Theorems of Pappus.
§ 110. The following general propositions concerning surfacesand solids of revolution, usually called Guldin’s theorems, are worth the reader’s attention.
If any plane curve revolve about any external axis situated in its plane, then
(a) the surface of the solid which is thereby generated is equal to the product of the perimeter of the revolving curve and the length of the path described by the centre of gravity of that perimeter;
(β) the volume of the solid is equal to the product of the area ol the revolving curve and the length of the path described by the centre of gravity of the revolving area.
We content ourselves with an example or two of the application of these theorems, referring to the article Infinitesimal Calculus for the proofs.
Example 1.—To find the surface and volume of a circular ring.— Let a be the distance of the centre of the generating curve, in this case a circle, from the axis of rotation, and r the radius of the circle, then perimeter of generating curve = 2τr, area of generating curve=πr 2 , and path described by the centre of gravity either of the perimeter or area = 2παj hence surface of ring=2τrr × 2τrα = 4π 2 ra, and volume of ring=πr 2 × 2πα = 2π 2 r¾.
Example 2. —To find the volume swept out by an ellipse whose axes are 2α and 2¼ revolving about an axis in its own plane whose distance from the centre of the ellipse is c.
Here area of generating curve = - ∣ rab, and path described by centre of gravity of area =2πc; hence volume generated =τrdb × 2πc=2π 2 abc. Example 3.—A circle of τ inches radius, with an inscribed regular hexagon, revolves about an axis a inches distant from its centre, and parallel to a side of the hexagon; to find the difference in area of the generated surfaces and volumes.
Here perimeter of circle = 2πr, and perimeter of hexagon=12 × τ sin 30° (§ 17)
= 6r; also area of circle=πr 2 , and area of hexagon = 3r 2 sin 6 0° (§ 18, 3)
= ⅜√3r 2 j hence difference of surfaces generated
= 4π 2 rα -12π∙αr=4παr(ιr - 3); and difference of volumes generated
= 2π 2 r 2 α-3ιrr 2 √3α
= πr⅛(2τr-3√3).
PART III. GAUGING.
§ 111. By gauging is meant the art of measuring the volume of a cask, or any portion of it. The subject is one of great interest and practical importance, but space will only permit us to discuss it very briefly. If the cask whose capacity we wish to determine be a solid of revolution, then its volume can at once be computed, either exactly or approximately, by the methods already described.
It is usual to divide casks into the following four classes according to the nature of the revolving curve :—
(a) the middle frustum of a spheroid;
(Ae) the middle frustum of a parabolic spindle;
(γ) two equal frusta of a paraboloid, united at their bases;
(δ) two equal frusta of a cone, united at their bases.
Casks of the second, third, and fourth variety are rarely met with in practice, and we shall accordingly confine our attention to the first kind, which is considered the true or model form of cask.
Let ABCD (fig. 56) be a section of the cask, and assume it to be the middle frustum of a prolate spheroid, then its volume = ⅛π(2δ 2 + Vfjk , where δ = OY, δ 1 =AX, and k=XX' (§ 99).
YY' is called the bung diameter, and AB or CD the head diameter.
An imperial gallon contains 277 ’ 274 cubic inches, and therefore the number of gallons in the above cask = τ (2⅜ 2 + ⅞!)⅛ π (2& 2 + b 2 }k 3x277’274 83Γ822 l + l '
^(¾⅛αX wherei - 2δ ∙ d '~ 2il '∙
whence we have the rule:—to the square of the head diameter add twice the square of the bung diameter, multiply the sum by the length and divide the result by 1059 Ί, and the answer is the content in imperial gallons.
Casks as ordinarily met with are not true spheroidal frusta, but it is better to consider them as such, calculate their capacity on this assumption, and then make allowance for the departure from the spheroidal form. The determination of the proper allowance to be made in each case is a matter depending on the skill and experience of the gauger, and proficiency in the art can only be attained by considerable practice.
§ 112. If the cask be very little curved, we obtain an approximation to its capacity by considering it as made up of two equal frusta of a cone, united at their bases. Hence from § 83 we have volume of cask = ⅛nA(r 2 + r 1 r 3 + r 2 ) nearly.
Here we neglect the small volumes generated by APY, YSD, BQY', and Y'RC; and therefore the volume is too small.
If we put r 1 r 3 =rj we obtain volume = ∖ κh{2r ∖ + rj), which is a little too large, and therefore the true volume lies between these two limits, and a very close approximation to it is said to be given by the formula
^{2rJ + r 2 -i(rJ-r≡)}.
§ 113. Ullage of a Cask.— The quantity of liquor contained in a cask partially filled and the capacity of the portion which is empty are termed respectively the wet and dry ullage.
(a) Ullage of a standing cask.— By means of the method applied in § 105, the following rule is deduced :—
Add the square of the diameter at the surface, the square of the diameter at the nearest end, and the square of double the diameter half-way between; multiply the sum by the length between the surface and the nearest end, and by Ό00472.
The product will be the wet or dry ullage according as the lesser portion of the cask is filled or empty.
(∕3) Ullage of a lying cask.-Υhe ullage in this case is found approximately on the assumption that it is proportional to the segment of the bung circle cut off by the surface of the liquor. The rule adopted in practice is ullage = t content × segmental area. (W. T.*)
